Is ${692297}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {692297}= &&{6}\cdot100000+ \\&&{9}\cdot10000+ \\&&{2}\cdot1000+ \\&&{2}\cdot100+ \\&&{9}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {692297}= &&{6}(99999+1)+ \\&&{9}(9999+1)+ \\&&{2}(999+1)+ \\&&{2}(99+1)+ \\&&{9}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {692297}= &&\gray{6\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {6}+{9}+{2}+{2}+{9}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${692297}$ is divisible by $9$ if ${ 6}+{9}+{2}+{2}+{9}+{7}$ is divisible by $9$ Add the digits of ${692297}$ $ {6}+{9}+{2}+{2}+{9}+{7} = {35} $ If ${35}$ is divisible by $9$ , then ${692297}$ must also be divisible by $9$ ${35}$ is not divisible by $9$, therefore ${692297}$ must not be divisible by $9$.